MCQ 511 Mark
If $x^2+ 5kx + 16 = 0,$ has equal roots, then the value of $k$ is :
- ✓
$\underline{+}\frac{8}{5}$
- B
$\underline{+}\frac{64}{25}$
- C
$\underline{+}\frac{25}{64}$
- D
$\underline{+}\frac{5}{8}$
AnswerCorrect option: A. $\underline{+}\frac{8}{5}$
Here $, a = 1, b = 5k, c = 16$
If $x^2+5 k x+16=0$ has equal roots
then, $b^2-4 a c=0$
$ \Rightarrow(5 k)^2-4 \times 1 \times 16=0$
$ \Rightarrow 25 k^2-64=0$
$ \Rightarrow 25 k^2=64$
$\Rightarrow \text{k}^{2} = \frac{64}{25}$
$\text{k} = \underline{+}\frac{8}{5}$
View full question & answer→MCQ 521 Mark
The values of $k$ for which the quadratic equation $2x^2- kx + k = 0$ has equal roots is :
- A
$8$ only
- B
$0$ only
- C
$4$
- ✓
$0, 8$
AnswerCorrect option: D. $0, 8$
If a quadratic equation $ax^2+ bx + c = 0, a \neq 0$ has two equal roots, then its discriminant value will be equal to zero
i.e. $D=b^2-4 a c=0$
Given, $2 x^2-k x+k=0$
For equal roots,
$ D=b^2-4 a c=0 $
$ \Rightarrow(-k)^2-4(2)^{(k)}=0 $
$ \Rightarrow k^2-8 k=0$
$\Rightarrow k(k - 8) = 0$
$\therefore k = 0, 8$
View full question & answer→MCQ 531 Mark
Choose the correct answer from the given four options in the following questions:
Which constant must be added and subtracted to solve the quadratic equation $9\text{x}^2+\frac{3}{4}\text{x}-\sqrt{2}=0$ by the method of completing the square ?
- A
$\frac{1}{8}$
- ✓
$\frac{1}{64}$
- C
$\frac{1}{4}$
- D
$\frac{9}{64}$
AnswerCorrect option: B. $\frac{1}{64}$
Given equation is $9\text{x}^2+\frac{3}{4}\text{x}-\sqrt{2}=0.$
$(3\text{x})^2+\frac{1}{4}(3\text{x})-\sqrt{2}=0$
On putting $3x = y,$ we have $\text{y}^2+\frac{1}{4}\text{y}-\sqrt{2}=0$
$\text{y}^2+\frac{1}{4}\text{y}+\Big(\frac{1}{8}\Big)^2-\Big(\frac{1}{8}\Big)^2-\sqrt{2}=0$
$\Big(\text{y}+\frac{1}{8}\Big)^2=\frac{1}{64}+\sqrt{2}$
$\Big(\text{y}+\frac{1}{8}\Big)^2=\frac{1+64\sqrt{2}}{64}$
Thus, $\frac{1}{64}$ must be added and subtracted to solve the given equation.
View full question & answer→MCQ 541 Mark
If the sum of the roots of the equation $k x^2+2 x+3 x=0$ is equal to their product, then the value of $k$ is :
- A
$\frac{1}{3}$
- B
$\frac{-1}{3}$
- C
$\frac{2}{3}$
- ✓
$\frac{-2}{3}$
AnswerCorrect option: D. $\frac{-2}{3}$
Given equation is $k x^2+2 x+3 x=0$
Comparing with $a x^2+b x+c=0,$
we have $a = k, b = 2, c = 3k$
Now,
Sum of the roots $=$ product of the roots
$\Rightarrow\frac{-2}{\text{k}}=\frac{\text{3k}}{\text{k}}$
$\Rightarrow\text{3k}=-2$
$\Rightarrow\text{k}=\frac{-2}3{}$
View full question & answer→MCQ 551 Mark
If one root of the equation $4\text{x}^2 - 2\text{x}+ (\lambda - 4) = 0 $ be the reciprocal of the other, then the value of $\lambda$ is :
AnswerLet one root be $\alpha$ then other root will be $\frac{1}{\alpha}$
Product of the root
$\Rightarrow\alpha \times \frac{1}{\alpha} = \frac{\text{c}}{\text{a}}$
$\Rightarrow 1=\frac{\lambda - 4}{4}$
$\Rightarrow\lambda - 4 = 4$
$\Rightarrow\lambda = 8$
View full question & answer→MCQ 561 Mark
The quadratic equation whose roots are $7 + \sqrt{3}$ and $7 - \sqrt{3}$ is :
- ✓
$ x^2-14 x+46=0 $
- B
$ x^2-14 x-46=0 $
- C
$ x^2+14 x+46=0 $
- D
$ x^2+14 x-46=0 $
AnswerCorrect option: A. $ x^2-14 x+46=0 $
Given : $\alpha = 7+\sqrt{3}$ and $\beta= 7-\sqrt{3}$
$\therefore\text{x}^{2}-(\alpha+\beta)\text{ x}+\alpha\beta = {0}$
$\Rightarrow\text{x}^{2} -(7+\sqrt{3}+7-\sqrt{3})\text{ x}+(7+\sqrt{3}) (7-\sqrt{3})=0$
$\Rightarrow\text{x}^{2} - {14}\text{x}+(49-3) = 0$
$\Rightarrow\text{x}^{2} - {14}\text{x}+46 = 0$
View full question & answer→MCQ 571 Mark
If the quadratic equation $kx(x - 2) + 6 = 0$ has equal roots, then the value of $k$ is :
AnswerGiven : $kx(x - 2) + 6 = 0$
$\Rightarrow k x^2-2 k x+6=0$
If the quadratic equation $k x^2-2 k x+6=0$ has equal roots,
then $b^2-4 a c=0 $
$\Rightarrow(-2 k)^2-4 \times k \times 6=0 $
$ \Rightarrow 4 k^2-24 k=0$
$\Rightarrow 4k(k - 6) = 0$
$\Rightarrow 4k = 0$ and $k - 6 = 0$
$\Rightarrow k = 0$ and $k = 6$
View full question & answer→MCQ 581 Mark
The quadratic equation $ax^2+ 2x + a = 0$ has two distinct roots, if :
AnswerCorrect option: C. $\text{a} = \underline{+} 1$
In the equation $a x^2+2 x+a=0$
$D=b^2-4 a c=(2)^2-4 \times a \times a=4-4 a^2$
Roots are real and equal
$ D=0 $
$ \Rightarrow 4-4 a^2=0$
$ \Rightarrow 4=4 a^2$
$ \Rightarrow 1=a^2 $
$ \Rightarrow a^2=1$
$\Rightarrow\text{a}^2 = (\underline+1)^2$
$\Rightarrow\text{ a} = \underline+1$
View full question & answer→MCQ 591 Mark
Choose the correct answer from the given four options in the following questions: Values of $k$ for which the quadratic equation $2x^2- kx + k = 0$ has equal roots is :
- A
$0$ only.
- B
$4.$
- C
$8$ only.
- ✓
$0, 8.$
AnswerCorrect option: D. $0, 8.$
Given equation is $2 x^2-k x+k=0$
On comparing with $a x^2+b x+c=0$, we get
$a = 2, b = -k$ and $c = k$
For equal roots, the discriminant must be zero.
i.e., $D=b^2-4 a c=0$
$\Rightarrow(-k)^2-4(2) k=0$
$\Rightarrow k^2-8 k=0$
$\Rightarrow k(k - 8) = 0$
$\therefore k = 0, 8$
Hence, the required values of $k$ are $0$ and $8$.
View full question & answer→MCQ 601 Mark
Choose the correct answer from the given four options in the following questions : The quadratic equation $2\text{x}^2-\sqrt{5}\text{x}+1=0$ has.
AnswerGiven equation is $2\text{x}^2-\sqrt{5}\text{x}+1=0$
On comparing with $a x^2+b x+c=0,$ we get
$a = 2, \text{b}=-\sqrt{5}$ and $c = 1$
$\therefore$ Discriminant, $D=b^2-4 a c$
$=\Big(-\sqrt{5}\Big)^2-4\times(2)\times(1)=5-8$
$=-3<0$
Since, discriminant is negative,
therefore quadratic equation $2\text{x}^2 - \sqrt{5}\text{x}+1=0$ has no real roots
i.e., imaginary roots.
View full question & answer→MCQ 611 Mark
The roots of the equation $2x^2- 6x + 3 = 0$ are :
- A
Real, unequal and rational.
- ✓
Real, unequal and irrational.
- C
- D
AnswerCorrect option: B. Real, unequal and irrational.
Given equation $2x^2- 6x + 3 = 0$
Here $,a = 2, b = -6, c = 3$
Discriminant $,D = b^2- 4ac$
$= (-6)^2- 4 \times 2 \times 3$
$= 36 - 24$
$= 12 < 0$
Also $,12$ is not a perfect square.
Hence, the roots of the given equation are real, unequal and irrational.
View full question & answer→MCQ 621 Mark
$x^2- 30x + 225 = 0$ have :
Answer$D = (-30)^2- 4 \times 1 \times 225$
$D = 900 - 900$
$D = 0.$
Real and Equal roots.
View full question & answer→MCQ 631 Mark
If one root of the equation $ax^2+ bx + c = 0$ is three times the other, then $b^2: ac =$
- A
$3 : 1$
- B
$3 : 16$
- ✓
$16 : 3$
- D
$16 : 1$
AnswerCorrect option: C. $16 : 3$
Quad equation is $ax^2+ bx + c = 0$
Let first root $=\alpha,$ Then
Second root $=3\alpha$
$\therefore$ Sum of root $=\alpha+3\alpha=\frac{-\text{b}}{\text{a}}$
$\Rightarrow4\alpha=\frac{-\text{b}}{\text{a}}$
$\Rightarrow\alpha=\frac{-\text{b}}{4\text{a}}\ ....(\text{i})$
and produt of roots $=\alpha\times3\alpha=\frac{\text{c}}{\text{a}}$
$\Rightarrow3\alpha^2=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha^2=\frac{\text{c}}{3\text{a}}$
$\Rightarrow\Big(\frac{-\text{b}}{4\text{a}}\Big)^2=\frac{\text{c}}{3\text{a}}\ [$From $(i)]$
$\Rightarrow\frac{\text{b}^2}{16\text{a}^2}=\frac{\text{c}}{3\text{a}} \ ($Dividing by $a)$
$\Rightarrow\frac{\text{b}^2}{16\text{a}}=\frac{\text{c}}{3}$
$\Rightarrow\frac{\text{b}^2}{\text{ac}}=\frac{16}{3}$
$\Rightarrow\text{b}^2:\text{ac}=16:3$
View full question & answer→MCQ 641 Mark
The roots of a quadratic equation $x^2- 4px + 4p^2- q^2= 0$ are :
- A
$2p + q, 2p + q$
- B
$p + 2q, p - 2q$
- C
$2p - q, 2p - q$
- ✓
$2p + q, 2p - q$
AnswerCorrect option: D. $2p + q, 2p - q$
Given : $x^2-4 p x+4 p^2-q^2=0$
$\Rightarrow(x-2 p)^2-q^2=0$
Using, ${a}^2-{b}^2=({a}+{b})({a}-{b})$,
$\Rightarrow (x - 2p + q) (x - 2p - q) = 0$
$\Rightarrow x - 2p + q = 0$ and $ x - 2p - q = 0$
$\Rightarrow x = 2p - q $ and $x = 2p + q$
View full question & answer→MCQ 651 Mark
If $p$ and $q$ are the roots of the equation $x^2+ px + q = 0,$ then :
- A
$p = -2, q = l$
- B
$p = -2, q = 0$
- ✓
$p = 1, q = -2$
- D
$b = 0, 9 = 1$
AnswerCorrect option: C. $p = 1, q = -2$
Given: sum of roots, $S = p + q = -p$ and product $pq = q$
$\Rightarrow q(p - 1) = 0$
i.e. $q = 0$ or $p = 1$
Now If $q = 0$ then $p = 0,$ this implies $p = q$
If $p = 1,$ then $p + q = -p$
$q = -2p$
$q = -2(1)$
$q = -2$
View full question & answer→MCQ 661 Mark
The common root of $2x^2+ x - 6 = 0$ and $x^2- 3x - 10 = 0$ is :
AnswerGiven : Putting $x = -2$ in given equations
$p(x)=2 x^2+x-6=0$ and $q(x)=x^2-3 x-10=0$
$\therefore p(-2)=2(-2)^3+(-2)-6=0$
$= 8 - 2 - 6 = 8 - 8 = 0$
$\therefore q(-2)=(-2)^2-3(-2)-10=0$
$= 4 + 6 - 10$
$= 10 - 10 = 0$
Since $,p(-2) = 0$ and $q(-2) = 0$
$\therefore -2$ is the common root of $2x^2+ x - 6 = 0$ and $x^2- 3x - 10 = 0.$
View full question & answer→MCQ 671 Mark
The roots of a quadratic equation are $5$ and $-2.$ Then, the equation is :
- A
$ x^2-3 x+10=0 $
- B
$ x^2+3 x-10=0 $
- C
$ x^2+3 x+10=0 $
- ✓
$ x^2-3 x-10=0 $
AnswerCorrect option: D. $ x^2-3 x-10=0 $
Sum of the roots $= 5 + (-2) = 3,$ product of roots $= 5 \times (-2) = -10.$
$\therefore x^2 - ($sum of roots$)\ x\ +$ product of roots $= 0.$
$ x^2-3 x-10=0 $
View full question & answer→MCQ 681 Mark
If the roots of the equations $(a^2+ b^2)x^2- 2b(a + c)x + (b^2+ c^2) = 0$ are equal, then :
AnswerCorrect option: B. $b^2= ac$
The given quadric equation is $\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$, and roots equal.
Here, $a=\left(a^2+b^2\right), b=-2 b(a+c)$ and $c=b^2+c^2$
As we know that $D=b^2-4 a c$
Putting the value of $a=\left(a^2+b^2\right), b=-2 b(a+c)$ and $c=b^2+c^2$
$ =\{-2 b(a+c)\}^2-4 \times\left(a^2+b^2\right) \times\left(b^2+c^2\right) $
$=4 a^2 b^2+4 b^2 c^2+8 a b^2 c-4\left(a^2 b^2+a^2 c^2+b^4+b^2 c^2\right) $
$ =4 a^2 b^2+4 b^2 c^2+8 a b^2 c-4 a^2 b^2-4 a^2 c^2-4 b^4-4 b^2 c^2 $
$ =+8 a b^2 c-4 a^2 c^2-4 b^4 $
$ =-4\left(a^2 c^2+b^4-2 a b^2 c\right) $
The given equation will have equal roots, if $D = 0$
$ -4\left(a^2 c^2+b^4-2 a b^2 c\right)=0 $
$ a^2 c^2+b^4-2 a b^2 c=0 $
$ \left(a c-b^2\right)^2=0 $
$ a c-b^2=0 $
$ a c=b^2 $
Thus, the correct answer is $(b)$
View full question & answer→MCQ 691 Mark
Choose the correct answer from the given four options in the following questions : Which of the following equations has the sum of its roots as $3?$
- A
$2\text{x}^2-3\text{x}+6=0.$
- ✓
$-\text{x}^2+3\text{x}-3=0.$
- C
$\sqrt{2}\text{x}^2-\frac{3}{\sqrt{2}}\text{x}+1=0.$
- D
$3\text{x}^2-3\text{x}+3=0$
AnswerCorrect option: B. $-\text{x}^2+3\text{x}-3=0.$
$(a)$ Given that, $2\text{x}^2-3\text{x}+6=0.$
On comparing with $ax^2+ bx + c = 0$, we get
$a = 2, b = -3$ and $c = 6$
$\therefore$ Sum of the roots $=\frac{-\text{b}}{\text{a}}=\frac{-(-3)}{2}=\frac{3}{2}$
So, sum of the roots of the quadratic equation $2\text{x}^2-3\text{x}+6=0.$ is not $3;$ so it is not the answer.
$(b) $ Given that, $-\text{x}^2+3\text{x}-3=0.$
On compare with $ax^2+ bx + c = 0$, we get
$a = -1, b = 3$ and $c = -3$
$\therefore$ Sum of the roots $=\frac{-\text{b}}{\text{a}}=\frac{-(3)}{-1}=3$
So, sum of the roots of the quadratic equation $-\text{x}^2+3\text{x}-3=0.$ is $3, $ so it is the answer.
$(c)$ Given that, $\sqrt{2}\text{x}^2-\frac{3}{\sqrt{2}}\text{x}+1=0$
$\Rightarrow\ 2\text{x}^2-3\text{x}+\sqrt{2}=0$
On comparing with $ax^2+ bx + c = 0,$ we get
$a = 2, b = -3$ and $\text{c}=\sqrt{2}$
$\therefore$ Sum of the roots $=\frac{-\text{b}}{\text{a}}=\frac{-(-3)}{2}=\frac{3}{2}$
So, sum of the roots of the quadratic equation $\sqrt{2}\text{x}^2-\frac{3}{\sqrt{2}}\text{x}+1=0$ is not $3,$ so it is not the answer,
$(d)$ Given that, $3x^2- 3x + 3 = 0$
$\Rightarrow x^2- x + 1 = 0$
On comparing with $ax^2+ bx + c = 0$, we get
$a = 1, b = -1$ and $c = 1$
$\therefore$ Sum of the roots $=\frac{-\text{b}}{\text{a}}=\frac{-(-1)}{1}=1$
So, sum of the roots of the quadratic equation $3x^2- 3x + 3 = 0$ is not $3,$ so it is not the answer.
View full question & answer→MCQ 701 Mark
If the sum and product of the roots of the equation $kx^2+ 6x + 4k = 0$ are equal, then $k =$
- A
$-\frac{3}{2}$
- ✓
$\frac{3}{2}$
- C
$\frac{2}{3}$
- D
$-\frac{2}{3}$
AnswerCorrect option: B. $\frac{3}{2}$
$k x^2+6 x+4 k=0$
Here $a = k, b = 6, c = 4k$
$\Rightarrow D=b^2-4 a c$
$\Rightarrow D=(6)^2-4 \times k \times 4 k$
$\Rightarrow D=36-16 k^2$
Roots are equal
$\Rightarrow D = 0$
$\Rightarrow 36-16 \mathrm{k}^2=0$
$\Rightarrow 16 \mathrm{k}^2=36$
$\Rightarrow\text{k}^2=\frac{36}{16}$
$\Rightarrow\text{k}^2=\Big(\frac{6}{4}\Big)^2$
$\Rightarrow\text{k}=\frac{6}{4}$
$\Rightarrow\text{k}=\frac{3}{2}$
View full question & answer→MCQ 711 Mark
If $a$ and $b$ are roots of the equation $x^2+ ax + b = 0,$ then $a + b =$
Answer$a$ and $b$ are the roots of the equation $x^2+ ax + b = 0$
Sum of roots $= -a$ and product of roots $= b$
Now $a + b = -a$
and $ab = b$
$\Rightarrow a = 1 ....(i)$
$\Rightarrow 2a + b = 0$
$\Rightarrow 2 \times 1 + b = 0$
$\Rightarrow b = -2$
Now $a + b = 1 - 2 = -1$
View full question & answer→MCQ 721 Mark
Choose the correct answer from the given four options in the following questions : Which of the following is not a quadratic equation?
- A
$ x^2-4 x+5=0 $
- B
$ x^2+3 x-12=0 $
- ✓
$ 2 x^2-7 x+6=0 $
- D
$ 3 x^2-6 x-2=0 $
AnswerCorrect option: C. $ 2 x^2-7 x+6=0 $
$(a)$ Substituting $x=2$ in $x^2-4 x+5$, we get
$(2)^2-4(2)+5$
$= 4 - 8 + 5 = 1 \neq 0.$
So $,x=2$ is not a root of $x^2-4 x+5=0$.
$(b)$ Substituting $x=2$ in $x^2+3 x-12$ we get
$(2)^2+3(2)-12$
$=4+6-12=-2\neq0$
So $, x = 2$ is not a root $x^2-3 x-12=0$.
$(c)$ Substituting $x=2$ in $2 x^2-7 x+6$, we get
$2(2)^2-7(2)+6=2(4)-14+6$
$= 8 - 14 + 6 = 14 - 14 = 0$
So $,x=2$ is root of the equation $2 x^2-7 x+6=0$.
$(d)$ Substituting $x=2$ in $3 x^2-6 x-2$, we get
$ 3(2)^2-6(2)-2 $
$ =12-12-2=-2 \neq 0$
So $,x=2$ is not a root of $3 x^2-6 x-2=0$.
View full question & answer→MCQ 731 Mark
If the sum of the roots of the equation $x^2-(k+6) x+2(2 k-1)=0$ is equal to half of their product, then $k =$
AnswerThe given quadric equation is $x^2-(k+6) x+2(2 k-1)=0,$ and roots are equal
Then find the value of $k$.
Let $\alpha$ and $\beta$ be two roots of given equation
And $, a = 1, b = -(k + 6)$ and of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$\alpha+\beta=\frac{-\{-(\text{k}+6)\}}{1}$
$\alpha+\beta=(\text{k}+6)$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\alpha\beta=\frac{2(2\text{k}-1)}{1}$
$\alpha\beta=2(2\text{k}-1)$
According to question, sum of the roots $=\frac{1}{2}\times$ product of the roots
$(\text{k}+6)=\frac{1}{2}\times2(2\text{k}-1)$
$\text{k}+6=2\text{k}-1$
$6+1=2\text{k}-\text{k}$
$7=\text{k}$
Therefore, the value of $k = 7$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 741 Mark
A quadratic equation whose one root is $3$ and the sum of the roots is zero is :
- A
$ x^2+9=0 $
- B
$ 9 x^2-1=0 $
- ✓
$ x^2-9=0 $
- D
$ 9 x^2+1=0 $
AnswerCorrect option: C. $ x^2-9=0 $
Let one root be $3$
then, other root will be $0 - 3 = -3$
$\therefore$ The quadratic equation will be $\text{x}^2 - (\alpha+\beta) + \alpha\beta = 0$
$\Rightarrow x^2- (3 - 3) x + 3 \times (-3) = 0$
$ \Rightarrow x^2-9=0 $
View full question & answer→MCQ 751 Mark
If the sum of the roots of the equation $\text{x}^2-\text{x}=\lambda(2\text{x}-1)$ is zero, then $\lambda=$
- A
$-2$
- B
$2$
- ✓
$-\frac{1}{2}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $-\frac{1}{2}$
$\Rightarrow\text{x}^2-\text{x}=\lambda(2\text{x}-1)$
$\Rightarrow\text{x}^2-\text{x}=2\lambda\text{x}-\lambda$
$\Rightarrow\text{x}^2-\text{x}-2\lambda\text{x}+\lambda=0$
$\Rightarrow\text{x}^2-(1+2\lambda)\text{x}+\lambda=0$
Sum of roots $=\frac{-\text{b}}{\text{a}}$
$=\frac{1+2\lambda}{1}$
$\frac{1+2\lambda}{1}=0$
$\Rightarrow2\lambda=-1$
$\lambda=-\frac{1}{2}$
View full question & answer→MCQ 761 Mark
If $\alpha$ and $\beta$ are the roots of the equation $3 x^2+8 x+2=0$ then $\Big(\frac{1}{\alpha}+\frac{1}{\beta}\Big)=?$
- A
$\frac{-3}{8}$
- B
$\frac{2}{3}$
- ✓
$-4$
- D
$4$
AnswerThe given equation is $3 x^2+8 x+2=0$
Here $, a = 3, b = 8, c = 2$
$\alpha+\beta=-\frac{8}{3}$ and $\alpha\times\beta=\frac{2}3{}$
Now, $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}$
$=\frac{\frac{-8}{3}}{\frac{2}{3}}=-4$
View full question & answer→MCQ 771 Mark
A quadratic equation whose one root is $2$ and the sum of whose roots is zero, is :
- A
$ x^2+a^4=0 $
- ✓
$ x^2-4=0 $
- C
$ 4 x^2-1=0 $
- D
$ x^2-2=0 $
AnswerCorrect option: B. $ x^2-4=0 $
Let $\alpha$ and $\beta$ be the roots of quadratic equation in such a way that $\alpha=2$
Then, according to question sum of the roots
$\alpha+\beta=0$
$2+\beta=0$
$\beta=-2$
And the product of the roots
$\alpha\cdot\beta=2\times(-2)$
$\alpha\cdot\beta=-4$
As we know that the quadratic equation
$\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta=0$
Putting the value of $\alpha$ and $\beta$ in above
Therefore, the require be
$\text{x}^2-0\times\text{x}+(-4)=0$
$\text{x}^2-4=0$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 781 Mark
The quadratic equation $\text{2x}^2 – \sqrt{5\text{x}} + 1 = 0$ has :
- A
More than $2$ real roots
- B
- C
- ✓
AnswerDiscriminant $= 5 - 4(2)(1) < 0$
$\therefore$ no real root.
View full question & answer→MCQ 791 Mark
The sum of the roots of the equation $x^2- 6x + 2 = 0$ is :
Answer$x^2- 6x + 2 = 0$
Comparing with $ax^2+ bx + c = 0,$
we have $a = 1, b = -6, c = 2$
Sum of the roots $=-\frac{\text{b}}{\text{a}}=-\frac{(-6)}{1}=6$
View full question & answer→MCQ 801 Mark
If one root of the equation $x^2+a x+3=0$ is $1,$ then its other root is :
AnswerThe quad equation is $x^2+a x+3=0$
One root $= 1$
and product of roots $=\frac{\text{c}}{\text{a}}=\frac{3}{1}=3$
Second root $=\frac{3}{1}=3$
View full question & answer→MCQ 811 Mark
$3x^2+ 2x - 1 = 0$ have :
Answer$D = b^2- 4ac$
$D = 2^2- 4 \times 3 \times (-1)$
$D = 4 + 12$
$D = 16$
$D > 0$.
Real and distinct roots.
View full question & answer→MCQ 821 Mark
In a cricket match, Kumble took three wickets less than twice the number of wickets taken by Srinath. The product of the number of wickets taken by these two is $20,$ then the number of wickets taken by Kumble is :
AnswerLet the number of wickets taken by Srinath be $x$
then, the number of wickets taken by Kumble will be $2x - 3$
According to question $, x(2x - 3) = 20$
$ \Rightarrow 2 x^2-3 x-20=0 $
$ \Rightarrow 2 x^2-8 x+5 x-20=0 $
$\Rightarrow 2x(x - 4) + 5(x - 4) = 0$
$\Rightarrow (x - 4) (2x + 5) = 0$
$\Rightarrow x - 4 = 0$ and $2x + 5 = 0$
$\Rightarrow \text{x = 4}$ and $\text{ x} =\frac{-5}{2}[\text{x}=\frac{-5}{2}$ is not possible$]$
The number of wickets taken by Srinath is $4$.
Then, the number of wickets taken by Kumble $= 2 \times 4 - 3 = 5$
View full question & answer→MCQ 831 Mark
$\sqrt{2}\text{x}^{2} - \text{3x}-5 = 0$ have :
Answer$\text{D} = (-3)^{2} - 4\times\sqrt{2}\times(-5)$
$\text{D} = 9+20\sqrt{2}$
$\text{D}>0.$
Real and distinct roots.
View full question & answer→MCQ 841 Mark
If $a$ and $b$ can take values $1, 2, 3, 4$. Then the number of the equations of the from $ax^2+ bx + 1 = 0$ having real roots is :
AnswerGiven that the equation $ax^2+ bx + 1 = 0$
For given equation to have real roots, discriminant $(\text{D})\geq0$
$\Rightarrow\text{b}^2-4\text{a}\geq0$
$\Rightarrow\text{b}^2\geq4\text{a}$
$\Rightarrow\text{b}\geq2\sqrt{\text{a}}$
Now, it is given that $a$ and $b$ can take values of $1, 2, 3$ and $4$
The above condition $\text{b}\geq2\sqrt{\text{a}}$ can be satisfied when
$b = 4$ and $a = 1, 2, 3, 4$
$b = 3$ and $a = 1, 2$
$b = 2$ and $a = 1$
So, there will be a maximum of $7$ equations for the values of $(a, b) $
$= (1, 4), (2, 4), (3, 4), (4, 4), (1, 3), (2, 3)$ and $(1, 2)$
Thus, the correct option is $(b)$
View full question & answer→MCQ 851 Mark
The discriminant of $4x^2+ 3x - 2 = 0$ is:
AnswerHere,
$a = 4, b = 3, c = -2$
Discriminant = $b^2- 4ac$
$= (3)^2- 4 \times 4 \times (-2)$
$= 9 + 32 = 41$
View full question & answer→MCQ 861 Mark
If $ax^2+ bx + c = 0$ has equal roots, then $c$ is equal to :
- A
$-\frac{\text{b}^2}{\text{2a}}$
- B
$-\frac{\text{b}^2}{\text{4a}}$
- C
$\frac{\text{b}^2}{\text{4a}}$
- ✓
$\frac{\text{b}^2}{\text{4a}}$
AnswerCorrect option: D. $\frac{\text{b}^2}{\text{4a}}$
If $ax^2+ bx + c = 0$ has equal roots,
then $b^2- 4ac = 0$
$\Rightarrow 4ac = b^2$
$\Rightarrow\text{c} = \frac{\text{b}^2}{\text{4a}}$
View full question & answer→MCQ 871 Mark
If the sum and product of the equations $k x^2+6 x+4 k=0$ are equal, then $k =$
- A
$\frac{-2}{3}$
- ✓
$\frac{-3}{2}$
- C
$\frac{2}{3}$
- D
$\frac{3}{2}$
AnswerCorrect option: B. $\frac{-3}{2}$
Given : $\alpha+\beta = \alpha\beta$
Sum of roots $=$ product of roots
$\Rightarrow\frac{-\text{b}}{\text{a}} = \frac{\text{c}}{\text{a}}$
$\Rightarrow-\text{b} = \text{c}$
$\Rightarrow -6 = \text{4k}$
$\Rightarrow\text{k}=\frac{-3}{2}$
View full question & answer→MCQ 881 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : The values of $x$ are $-a^2,$ a for a quadratic equation $2 \times 2+a x-a^2=0$.
Reason : For quadratic equation $a \times 2+b x+c=0 \text{x}=\frac{-\text{b}\pm\sqrt{\text{b}^2-4\text{ac}}}{2\text{a}}$
- A
If both assertion and reason are true and reason is the correct explanation of assertion.
- B
If both assertion and reason are true but reason is not the correct explanation of assertion.
- C
If assertion is true but reason is false.
- ✓
If both assertion and reason are false.
AnswerCorrect option: D. If both assertion and reason are false.
$=2\text{x}^2+\text{ax}−\text{a}^2=0$
$=\text{x}=\frac{-\text{a}\pm\sqrt{\text{a}^2-8\text{a}^2}}{4}$
$\Rightarrow\text{x}=\frac{-\text{a}+3\text{a}}{4}$
$\Rightarrow\text{x}=\frac{2\text{a}}4,\frac{-4\text{a}}{4}$
$\Rightarrow\text{x}=\frac{\text{a}}2,-\text{a}$
View full question & answer→MCQ 891 Mark
For what values of $k,$ the equation $k x^2-6 x-2=0$ has real roots ?
AnswerCorrect option: B. $\text{k}\ge\frac{-9}{2}$
Given, the roots of $k x^2-6 x-2=0$ are real
$\Rightarrow\text{D}\ge0$
$\Rightarrow\text{b}^2-\text{4ac}\ge0$
$\Rightarrow(-6)^2-4\times\text{k}\times(-2)\ge0$
$\Rightarrow36+\text{8k}\ge0$
$\Rightarrow\text{8k}\ge-36$
$\Rightarrow\text{k}\ge-\frac{9}{2}$
View full question & answer→MCQ 901 Mark
If $x = 1$ is a common root of the equations $ax^2+ ax + 3 = 0$ and $x^2+ x + b = 0,$ then $ab =$
AnswerIn the equation $a^2+a x+3=0$ and $x^2+x+b=0$
Substituting the value of $x = 1,$ then in $a x^2+a x+3=0$
$a(1)^2+a(1)+3=0$
$\Rightarrow a + a + 3 = 0$
$\Rightarrow 2a + 3 = 0$
$\Rightarrow 2a = -3$
$\Rightarrow a = -32$
and in $x^2+x+b=0$
$(1)^2+ 1 + b = 0$
$\Rightarrow 1 + 1 + b = 0$
$\Rightarrow b = -2$
$\therefore\text{ab} = \frac{-3}{2}\times(-2) = 3$
View full question & answer→MCQ 911 Mark
$(x - 1) (2x - 1) = 0$ discriminant of the given equation is :
Answer$(x - 1) (2x - 1) = 0$
$ 2 x^2-3 x+1=0 $
$ D=b^2-4 a c $
$ D=(-3)^2-4 \times 2 \times 1$
$D = 9 - 8$
$D = 1$
View full question & answer→MCQ 921 Mark
If $p$ and $q$ are the roots of the equation $x^2- px + q + 0,$ then :
- ✓
$p = 1, q = -2$
- B
$p = 0, q = 1$
- C
$p = -2, q = 0$
- D
$p = -2, q = 1$
AnswerCorrect option: A. $p = 1, q = -2$
Given that $p$ and $q$ be the roots of the equation $x^2- Px + q + 0$
Then find the value of $p$ and $q.$
Here $, a = 1, b = -p$ and $c = q$
$p$ and $q$ be the roots of the given equation
Therefore, sum of the roots
$\text{p}+\text{q}=\frac{-\text{b}}{\text{a}}$
$\text{p}+\text{q}=\frac{-\text{p}}{1}$
$\text{p}+\text{q}=-\text{p}$
$\text{q}=-\text{p}-\text{p}$
$\text{q}=-2\text{p}\ ....(\text{i})$
Product of the roots
$\text{p}\times\text{q}=\frac{\text{q}}{1}$
As we know that
$\text{p}=\frac{\text{q}}{\text{q}}$
$\text{p}=1$
Putting the value of $p = 1$ in equation $(i)$
$q = -2 \times 1$
$q = -2$
Therefore, the value of $p = 1, q = -2$
Thus, the correct answer is $(a)$
View full question & answer→MCQ 931 Mark
If one root of the equation $3x^2- 10x + 3 = 0$ is $\frac{1}{3}$ then the other root is :
- A
$\frac{-1}{3}$
- B
$\frac{1}{3}$
- C
$-3$
- ✓
$3$
AnswerLet the other root be $\alpha.$
Given equation is $3 x^2-10 x+3=0$
Comparing with $ax^2+bx+c=0,$
we have $a = 3, b = -10, c = 3$
Product of the roots $=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha\times\frac{1}{3}=\frac{3}{3}$
$\Rightarrow\alpha=3$
View full question & answer→MCQ 941 Mark
If the equation $(a^2+ b^2)x^2- 2(ac + bd)x + c^2+ d^2 = 0$ has equal roots, then :
AnswerCorrect option: B. $\text{ad}=\text{bc}$
In the equation
$ \Rightarrow\left(a^2+b^2\right) x^2-2(a c+b d) x+\left(c^2+d^2\right)=0 $
$ \Rightarrow D=B^2-4 \ AC $
$ \Rightarrow D=[-2(a c+b d)]^2-4\left(a^2+b^2\right)\left(c^2+d^2\right) $
$ \Rightarrow D=4\left[a^2 c^2+b 2 d^2+2 a b c d\right]-4\left[a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right] $
$ \Rightarrow D=4 a^2 c^2+4 b^2 d^2+8 a b c d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 b^2 d^2 $
$ \Rightarrow D=8 a b c d-4 a^2 d 2-4 b^2 c^2 $
$ \Rightarrow D=-4\left[a^2 d^2+b^2 c^2-2 a b c d\right] $
$ \Rightarrow D=-4(a d-b c)^2 $
$\because$ Roots are equal
$\therefore D = 0$
$\Rightarrow -4(ad - bc)^2= 0$
$\Rightarrow ad - bc = 0$
$\Rightarrow ad = bc$
View full question & answer→MCQ 951 Mark
The length of a rectangular field exceeds its breadth by $8m$ and the area of the field is $240m^2$. The breadth of the field is :
AnswerLet the breadth of the rectangle be $x m$.
Then, length of the rectangle $= (x + 8)m$
Now Area $= 240m^2$
$\Rightarrow$ Length $\times$ Breadth $= 240$
$\Rightarrow x(x + 8) = 240$
$\Rightarrow x^2+8 x=240$
$\Rightarrow x^2+8 x-240=0$
$\Rightarrow x^2+20 x-12 x-240=0$
$\Rightarrow x(x + 20) - 12(x + 20) = 0$
$\Rightarrow (x + 20)(x - 12) = 0$
$\Rightarrow x + 20 = 0$ or $x - 12 = 0$
$\Rightarrow x = -20$ or $x = 12$
$\Rightarrow x = 12 \ ($Breadth cannot be negative$)$.
View full question & answer→MCQ 961 Mark
If $\alpha$ and $\beta$ are the roots of $a x^2+b x+c=0,$ then the wrong statement is :
- A
$\frac{1}{\alpha}+\frac{1}{\beta} = \frac{-\text{b}}{\text{c}}$
- B
$\alpha\beta = \frac{\text{c}}{\text{a}}$
- C
$\alpha^2+\beta^2=\frac{\text{b}^2-2\text{ac}}{\text{a}^2}$
- ✓
$\alpha+\beta=\frac{\text{b}}{\text{a}}$
AnswerCorrect option: D. $\alpha+\beta=\frac{\text{b}}{\text{a}}$
If $\alpha$ and $\beta$ are the roots of $a x^2+b x+c=0,$
then $\alpha+\beta=\frac{\text{b}}{\text{a}}$
View full question & answer→MCQ 971 Mark
The roots of the equation $a x^2+b x+c=0$ will be reciprocal of each other if :
AnswerCorrect option: C. $c = a$
Product of the roots $=\frac{\text{c}}{\text{a}}$
Also $,{\alpha}\times\frac{1}{\alpha}=1$
$\Rightarrow\frac{\text{c}}{\text{a}}=1$
$\Rightarrow\text{c}=\text{a}$
View full question & answer→MCQ 981 Mark
The smallest value of $k$ for which the quadratic equation $x^2+ kx + 9 = 0$ has real roots is :
AnswerIf the quadratic equation $x^2+ kx + 9 = 0$ has real roots
then $, b^2-4 a c>0 $
$ \Rightarrow k^2-4 \times 1 \times 9>0 $
$ \Rightarrow k^2-36>0 $
$ \Rightarrow k^2>36 $
$\Rightarrow\text{k} > \underline{+ }{ 6}$
$\therefore$ The smallest value of $k$ is $-6$.
View full question & answer→MCQ 991 Mark
A two $-$ digit number is such that the product of the digits is $20$. When $9$ is added to the number then the digits interchange their places. The number is :
View full question & answer→MCQ 1001 Mark
The sum of a number and its reciprocal is $2\frac{1}{20}.$ The number is :
- ✓
$\frac{5}{4}$ or $\frac{4}{5}$
- B
$\frac{4}{3}$ or $\frac{3}{4}$
- C
$\frac{5}{6}$ or $\frac{6}{5}$
- D
$\frac{1}{6}$ or $6$
AnswerCorrect option: A. $\frac{5}{4}$ or $\frac{4}{5}$
Let the number be $x.$
Then, $\text{x}+\frac{1}{\text{x}}=\frac{41}{20}$
$\Rightarrow\frac{\text{x}^2+1}{\text{x}}=\frac{41}{20}$
$ \Rightarrow 20 x^2+20=41 x $
$ \Rightarrow 20 x^2-41 x+20=0 $
$ \Rightarrow 20 x^2-25 x-16 x+20=0 $
$\Rightarrow 5x(4x - 5) - 4(4x - 5) = 0$
$\Rightarrow (4x - 5)(5x - 4) = 0$
$\Rightarrow 4x - 5 = 0$ or $5x - 4 = 0$
$\Rightarrow 4x = 5$ or $5x = 4$
$\Rightarrow\text{x}=\frac{5}4{}$ or $\text{x}=\frac{4}{5}$
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